-11
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You have 6 boxes. You can use the digits from 1 to 9 but not 0. Digit repetition is not allowed. The total sum of the numbers/digits should be 20.

$$ \huge\square\square\square\\ \huge\square\square\square $$

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9
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It's not possible. The sum of the 6 smallest digits (1 through 6) is 21, already greater than 20.

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  • $\begingroup$ So, are we allowed to use negative numbers to solve this? $\endgroup$ – Mehmood Ali Nov 4 '15 at 17:29
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    $\begingroup$ The question is yours, how am I supposed to know. Haha. $\endgroup$ – Fimpellizieri Nov 4 '15 at 18:01
  • $\begingroup$ Yeah, it is my question but can we use negative numbers to solve this according to mathematics rules? Is that possible? $\endgroup$ – Mehmood Ali Nov 4 '15 at 18:04
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    $\begingroup$ Digits are not negative or positive, I am not sure I understand you. $\endgroup$ – Fimpellizieri Nov 4 '15 at 18:05
  • $\begingroup$ Alright, I got you. $\endgroup$ – Mehmood Ali Nov 4 '15 at 18:14
9
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Seems pretty easy to me.

$$ \huge[9][1][8]\\ \huge[2]\square\square $$

The rules don't say I need to fill in all of the boxes.

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3
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If we add the tag, I have a solution:

Write 4, 5, and 8 in the first three boxes
Cut off three of your fingers (also called "digits") and place one in each of the other three boxes.
Now you have 6 digits (three numbers, three fingers) in the boxes
Each finger has a value of 1 so the sum is $1+1+1+4+5+8=20$

Alternatively, if fingers have a value of 0, then us the numbers 2, 3, 4, 5, 6 and only cut off one finger (value of 0). Now the sum is $0+2+3+4+5+6=20$

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  • $\begingroup$ That's pretty bloody. I'd rather skip the puzzle than lose three fingers. $\endgroup$ – Illyasviel Nov 5 '15 at 15:38
3
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Inspired by @AlbertRenshaw's post:

Use hexadecimal. Then $2 + 4 + 5 + 6 + 7 + 8 = 20_{16}\;(=32_{10})$.

This can actually work with any base from 11 to 19:

$1 + 2 + 3 + 4 + 5 + 7 = 20_{11}\;(=22_{10})$
$1 + 2 + 3 + 4 + 5 + 9 = 20_{12}\;(=24_{10})$
$1 + 2 + 3 + 4 + 7 + 9 = 20_{13}\;(=26_{10})$
$1 + 2 + 3 + 6 + 7 + 9 = 20_{14}\;(=28_{10})$
$1 + 2 + 3 + 7 + 8 + 9 = 20_{15}\;(=30_{10})$
$1 + 4 + 5 + 7 + 8 + 9 = 20_{17}\;(=34_{10})$
$1 + 5 + 6 + 7 + 8 + 9 = 20_{18}\;(=36_{10})$
$3 + 5 + 6 + 7 + 8 + 9 = 20_{19}\;(=38_{10})$

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0
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1,2,3,4,5,7

And the puzzle is in base 9.

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  • 1
    $\begingroup$ Alas, if the puzzle were in base 9, 9 would not be a valid digit. $\endgroup$ – Ian MacDonald Nov 5 '15 at 15:03
  • $\begingroup$ @IanMacDonald 9 = ∞ in base 9. xD $\endgroup$ – Albert Renshaw Nov 5 '15 at 16:54
  • $\begingroup$ That's not quite how math works. :P $\endgroup$ – Ian MacDonald Nov 5 '15 at 17:05

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